Question

Questions: Figure 27-64 shows the circuit of a flashing lamp, like those attached to barrels at highway construction sites. The fluorescent lampL(of negligible capacitance) is connected in parallel across the capacitor C of an RC circuit. There is a current through the lamp only when the potential difference across it reaches the breakdown voltage${\mathit{V}}_{\mathbf{L}}\mathbf{=}\mathbf{72}\mathbf{.}\mathbf{0}\mathbf{\text{V}}$ ; then the capacitor discharges completely through the lamp and the lamp flashes briefly. For a lamp with breakdown voltage wired to an 0.950 V ideal battery and a $\mathbf{0}\mathbf{.}\mathbf{150}\mathit{\mu }\mathit{F}$ capacitor, what resistance Ris needed for two flashes per second?

Short answer

Answer:

The resistance R needed for two flashes per second is $2.35\times {10}^6\Omega$.

Step-by-step solution

The given data

The breakdown voltage of the circuit,${\text{V}}_{\text{L}}=72.0\text{V}$

Emf of the ideal battery, role="math" localid="1663146516011" $\epsilon =95.0V$

The capacitance of the capacitor,$0.150\mu \text{F}$

Understanding the concept of breakdown voltage

The threshold voltage where the breakdown of the circuit initiates is called the breakdown voltage of the circuit. The minimum voltage at which an insulator experiences momentary conduction is the breakdown voltage. Using the potential difference concept, we can get the resistance of the resistor in the given circuit.

Formula:

The potential difference of an RC circuit, $V=V_0\left(1-e^{-t/RC}\right)$ (1)

a) Calculation of the resistance for two flashes per second

The time it takes for the voltage difference across the capacitor to reach V_L can be given using equation (1) as follows:

$V_L=\epsilon \left(1-e^{-t/RC}\right).............................\left(2\right)$

Using the given data in equation (2), the resistance needed for two flashes per second can be given as follows:

$$\begin{gathered} \left(\text{Given, two flashes per second, 1 flash = 1/2 s or 0.5s}\right) \\ R=\frac{t}{C\ln \epsilon /\left(\epsilon -V_L\right)} \\ =\frac{\left(0.5\text{s}\right)}{\left(0.150\times {10}^{-6}\text{F}\right)\ln \left(\left(95.0\text{V}\right)/\left(95\text{V}-72\text{V}\right)\right)} \\ =2.35\times {10}^6\Omega \end{gathered}$$

Hence, the value of the resistance is $2.35\times {10}^6\Omega$